Question

If $\cos 50 = a$ , then how do you express $\tan 130$ in terms of $a$ ?

Answer 1

Hint : To solve this problem we have to start with the formula of $\tan {130^ \circ } = \dfrac{{\sin {{130}^ \circ }}}{{\cos {{130}^ \circ }}}$ and we have to equal the value of $\sin {130^ \circ }$ and $\cos {130^ \circ }$ to $\cos {50^ \circ }$ , so that we can change it in term of $a$ . And then if we substitute the value we get our required answer.

Complete step-by-step answer :
Let us consider the given question,
$\cos {50^ \circ } = a$
We know that $\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}$ , let us take $\theta = {130^ \circ }$ as given in the question and it becomes,
$\tan {130^ \circ } = \dfrac{{\sin {{130}^ \circ }}}{{\cos {{130}^ \circ }}}$
First let us define $\sin {130^ \circ }$ , ${130^ \circ }$ can be written as ${90^ \circ } + {40^ \circ }$,
$\sin {130^ \circ } = \sin ({90^ \circ } + {40^ \circ })$
Now consider the formula $\sin (A + B) = \sin A\cos B + \cos A\sin B$ where $A = {90^ \circ }$ and $b = {40^ \circ }$ substituting we get,
$\sin ({90^ \circ } + {40^ \circ }) = \sin {90^ \circ }\cos {40^ \circ } + \cos {90^ \circ }\sin {40^ \circ }$
The value for $\sin {90^ \circ } = 1$ and $\cos {90^ \circ } = 0$ , the above equation becomes,
$\sin ({90^ \circ } + {40^ \circ }) = (1)\cos {40^ \circ } + (0)\sin {40^ \circ } \\\ \sin ({90^ \circ } + {40^ \circ }) = \cos {40^ \circ } \\\$
Step: -2
Here, the $\theta = {130^ \circ }$ can also be written as ${180^ \circ } - {50^ \circ }$ and
$\sin ({130^ \circ }) = \sin ({180^{^ \circ }} - {50^{^ \circ }})$
The formula for $\sin (A + B) = \sin A\cos B - \cos A\sin B$ , the value of $A = {180^ \circ }$ and $B = {50^ \circ }$ , substiting the values we get,
$\sin ({180^{^ \circ }} - {50^{^ \circ }}) = \sin {180^{^ \circ }}\cos {50^{^ \circ }} - \sin {50^{^ \circ }}\cos {180^{^ \circ }}$
The value for $\sin {180^{^ \circ }} = 0$ and $\cos {180^{^ \circ }} = - 1$ ,
$\sin ({180^{^ \circ }} - {50^{^ \circ }}) = (0)\cos {50^{^ \circ }} - \sin {50^{^ \circ }}( - 1)$
$\sin ({180^{^ \circ }} - {50^{^ \circ }}) = \sin {50^ \circ }$
And therefore, we can say,
$\cos {40^ \circ } = \sin {50^ \circ }$ and therefore,
$\sin {130^ \circ } = \sin {50^ \circ }$
Squaring on both sides, we get,
${\sin ^2}{130^ \circ } = {\sin ^2}{50^ \circ }$
We know that ${\sin ^2}\theta = 1 - {\cos ^2}\theta$ ,
$\sin {130^ \circ } = \sqrt {1 - {{\cos }^2}130}$
From the question we know that,
$\cos {50^ \circ } = a$ and by substituting the value we get,
$\sin {130^ \circ } = \sqrt {1 - {a^2}}$ … (1)
Step: -3
Similarly,
$\cos {130^ \circ } = \cos ({90^ \circ } + {40^ \circ })$
Now consider the formula $\cos (A + B) = \cos A\cos B - \sin A\sin B$ where $A = {90^ \circ }$ and $b = {40^ \circ }$ substituting we get,
$\cos ({90^ \circ } + {40^ \circ }) = \cos {90^ \circ }\cos {40^ \circ } + \sin {90^ \circ }\sin {40^ \circ }$
The value for $\sin {90^ \circ } = 1$ and $\cos {90^ \circ } = 0$ , the above equation becomes,
$\cos ({90^ \circ } + {40^ \circ }) = (0)\cos {40^ \circ } + (1)\sin {40^ \circ } \\\ \cos ({90^ \circ } + {40^ \circ }) = \sin {40^ \circ } \\\$
Step: - 4
Here, the $\theta = {130^ \circ }$ can also be written as ${180^ \circ } - {50^ \circ }$ and
$\cos ({130^ \circ }) = \cos ({180^{^ \circ }} - {50^{^ \circ }})$
The formula for $\cos (A - B) = \cos A\cos B + \sin A\sin B$ , the value of $A = {180^ \circ }$ and $B = {50^ \circ }$ , substituting the values we get,
$\cos ({180^{^ \circ }} - {50^{^ \circ }}) = \cos {180^{^ \circ }}\cos {50^{^ \circ }} + \sin {50^{^ \circ }}\sin {180^{^ \circ }}$
The value for $\sin {180^{^ \circ }} = 0$ and $\cos {180^{^ \circ }} = - 1$ $\cos {180^{^ \circ }} = - 1$
$\cos ({180^{^ \circ }} - {50^{^ \circ }}) = ( - 1)\cos {50^{^ \circ }} - \sin {50^{^ \circ }}(0)$
$\cos ({180^{^ \circ }} - {50^{^ \circ }}) = - \cos {50^ \circ }$
And therefore, we can say,
$\sin {40^ \circ } = - \cos {50^ \circ }$ and therefore,
$\cos {130^ \circ } = - \cos {50^ \circ }$
From the question we know that,
$\cos {50^ \circ } = a$ and by substituting the value we get,
$\cos {130^ \circ } = - a$ … (2)
Step: -5
Substituting (1) and (2) in $\tan {130^ \circ } = \dfrac{{\sin {{130}^ \circ }}}{{\cos {{130}^ \circ }}}$ , we get
$\tan {130^ \circ } = - \dfrac{{\sqrt {1 - {a^2}} }}{a}$
This is our required answer.
So, the correct answer is “$- \dfrac{{\sqrt {1 - {a^2}} }}{a}$”.

Note : While dividing the $\theta$ either we can perform addition or subtraction, both the ways were accepted, for which the formula should be convenient to the given problems. We can also use the both in many cases.